Question

If \(3\sin^4 x + 2\cos^4 x = \frac{6}{5}\) and \(x\) is an acute angle, then \(\tan 2x =\)

• A. \( \dfrac{2\sqrt{6}}{5} \)
• B. \( 2\sqrt{6} \)
• C. \( \dfrac{3\sqrt{2}}{5} \)
• D. \( \dfrac{2\sqrt{3}}{5} \)

Hint:

Use substitution for powers of sine and cosine, and remember the identity for double angle: \[ \tan 2x = \frac{2 \tan x}{1 - \tan^2 x} \]

Answer 1

The Correct Option is B

Step 1: Let \(\sin^2 x = a\), then \(\cos^2 x = 1 - a\).
\[ \sin^4 x = a^2,\quad \cos^4 x = (1 - a)^2 \] Substitute into the given equation: \[ 3a^2 + 2(1 - a)^2 = \frac{6}{5} \]
Step 2: Expand and simplify. \[ 3a^2 + 2(1 - 2a + a^2) = \frac{6}{5} \] \[ 3a^2 + 2 - 4a + 2a^2 = \frac{6}{5} \] \[ 5a^2 - 4a + 2 = \frac{6}{5} \]
Step 3: Multiply through by 5 to eliminate denominator.
\[ 25a^2 - 20a + 10 = 6 \quad \Rightarrow \quad 25a^2 - 20a + 4 = 0 \]
Step 4: Solve the quadratic equation.
\[ a = \frac{20 \pm \sqrt{(-20)^2 - 4 \cdot 25 \cdot 4}}{2 \cdot 25} = \frac{20 \pm \sqrt{400 - 400}}{50} = \frac{20}{50} = \frac{2}{5} \] So, \(\sin^2 x = \frac{2}{5}\), and since \(x\) is acute, \(\sin x = \sqrt{\frac{2}{5}}\), \(\cos x = \sqrt{1 - \frac{2}{5}} = \sqrt{\frac{3}{5}}\)
Step 5: Use identity for \(\tan 2x\):
\[ \tan 2x = \frac{2 \tan x}{1 - \tan^2 x},\quad \tan x = \frac{\sin x}{\cos x} = \sqrt{\frac{2}{3}} \]
Step 6: Substitute in the identity.
\[ \tan 2x = \frac{2 \cdot \sqrt{\frac{2}{3}}}{1 - \frac{2}{3}} = \frac{2 \cdot \sqrt{\frac{2}{3}}}{\frac{1}{3}} = 6 \cdot \sqrt{\frac{2}{3}} = \sqrt{6} \cdot 2 = 2\sqrt{6} \]